Question

When $Cu ^{2+}$ ion is treated with $KI$, a white precipitate, $X$ appears in solution The solution is titrated with sodium thiosulphate, the compound $Y$ is formed $X$ and $Y$ respectively are

• A. $X = CuI _2 \,\,\,Y = Na _2 S _2 O _3$
• B. $X = CuI _2 \,\,\, Y = Na _2 S _4 O _6$
• C. $X = Cu _2 I _2 \,\,\,Y = Na _2 S _4 O _6$
• D. $X = Cu _2 I _2 \,\,\, Y = Na _2 S _4 O _5$

Answer 1

Answer: (C)

$X = Cu _2 I _2 \,\,\,Y = Na _2 S _4 O _6$

Answer 2

Cu2++2KI⟶ Unstable CuI2​​↓+2K+
I−is strong R.A it reduces Cu2+ to Cu+
2CuI2​⟶ (White)’X’ Cu2​I2​↓​+I2​
KI+I2​⟶K+I3−​(Brown solution)
I3−​⇌I2​+I−
KI3​+Na2​S2​O3​→KI+(Y)Na2​S4​O6​​